You have found the following ages (in years) of 6 zebras. Those zebras were randomly selected from the 25 zebras at your local zoo: $ 3,\enspace 13,\enspace 19,\enspace 27,\enspace 4,\enspace 13$ Based on your sample, what is the average age of the zebras? What is the standard deviation? You may round your answers to the nearest tenth.
Explanation: Because we only have data for a small sample of the 25 zebras, we are only able to estimate the population mean and standard deviation by finding the sample mean $({\overline{x}})$ and sample standard deviation $({s})$ To find the sample mean , add up the values of all $6$ samples and divide by $6$ $ {\overline{x}} = \dfrac{\sum\limits_{i=1}^{{n}} x_i}{{n}} = \dfrac{\sum\limits_{i=1}^{{6}} x_i}{{6}} $ $ {\overline{x}} = \dfrac{3 + 13 + 19 + 27 + 4 + 13}{{6}} = {13.2\text{ years old}} $ Find the squared deviations from the mean for each sample. Since we don't know the population mean, estimate the mean by using the sample mean we just calculated {104.04} + {0.04} + {33.64} + {190.44} + {84.64} + {0.04}} {{6 - 1}} $ {s^2} = \dfrac{{412.84}}{{5}} = {82.57\text{ years}^2} $ As you might guess from the notation, the sample standard deviation $({s})$ is found by taking the square root of the sample variance $({s^2})$ ${s} = \sqrt{{s^2}}$ $ {s} = \sqrt{{82.57\text{ years}^2}} = {9.1\text{ years}} $ We can estimate that the average zebra at the zoo is 13.2 years old. There is also a standard deviation of 9.1 years.